# What is the empirical formula of a phosphoric acid that contains 0.3086 g of hydrogen, 3.161 g of phosphorus, and 6.531 g of oxygen?

Apr 29, 2016

${H}_{2}$ ${P}_{4}$ ${O}_{1}$

#### Explanation:

In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound. On decomposition , we obtained 0.3086 g of Hydrogen , 3.161 g of Phosphorus, and 6.531 g of Oxygen.

*Step 1* Calculating the number of moles of different elements

n_$H$= mass of H/ molar mass of H

n_$H$= 0.3086 g / 1 g $m o {l}^{- 1}$

n_$H$= 0.3086 mol

n_P = mass of P/ molar mass of P

n_P$= 3.161 \frac{g}{31} g$mol^(-1)

n_$P$= 1.02 mole

n_$O$ = mass of O / molar mass of O

n_$O$= 6.531 g / 32 g $m o {l}^{- 1}$

n_#$O$= 0.2 mol

Calculating the Simple Ratio of the moles

H= 0.3806/ 0.2 =4 , P = 1.02 / 0.2 =5 , O= 0.2/0.2 =1

${H}_{2}$ ${P}_{4}$ ${O}_{1}$